Peckness of Edge Posets

0.26 MB·English

by  David Hemminger

PECKNESS OF EDGE POSETS
DAVIDHEMMINGER,AARONLANDESMAN,ZIJIAN YAO
Abstract. ForanygradedposetP,wedefineanewgradedposet,E(P),whoseelementsaretheedges
5 in the Hasse diagram of P. For any group, G, acting on the boolean algebra, Bn, we conjecture that
1 E(Bn/G) is Peck. We prove that the conjecture holds for “common cover transitive” actions. We give
0 some infinite families of common cover transitive actions and show that the common cover transitive
2 actionsareclosedunderdirectandsemidirectproducts.

n
a
J

9 1. Introduction
1
Let P be a finite graded poset of rank n. In this paper we study the structure of the edges in the

] HassediagramofP. Tothis end,wedefineanendofunctorEonthecategoryoffinite gradedposetswith
O
rank-preserving morphisms as follows
C
. Definition 1.1. For P the category of graded posets, define the functor of edges E: P → P as follows.
h Given P ∈ P, the elements of the graded poset E(P) are pairs (x,y) where x,y ∈ P, x ≤ y, and
t P
a rk(y) = rk(x)+1. Define the covering relation ⋖E on E(P) by (x,y)⋖E(x′,y′) if x⋖P x′ and y⋖P y′.
m Then define the relation ≤E on E(P) to be the transitive closure of ⋖E.
[ LetQ be a finite gradedposet ofrank n. Givena morphismf: P →Q, define E(f): E(P)→E(Q) by

E(f)(x,y)=(f(x),f(y)).
1
v We will show that E(P) is a well-defined graded poset in Section 3. Note that an edge in the Hasse
0
diagram of P can be identified with a pair (x,y) ∈ P ×P such that x⋖y, and the edges in the Hasse
4
5 diagram are in bijection with elements (x,y) ∈ E(P) via this identification. With this in mind, we will
4 frequently refer to E(P) as the edge poset of P.
0
. Example 1.2. We give an example of an edge poset in Figure 1. Note that it is important we declare
1
0 the relation ≤E to be the transitive closure of ⋖E. If instead we defined a relation ≤E′ on E(P) by
5 (x,y) ≤ (a,b) if x ≤ a,y ≤ b, then E(P) would not necessarily be a graded poset. In Figure 1 we give
1 an example of a poset P for which E(P) is not graded with the relation ≤E′. In Figure 1 it is clear that
: E(P) is a graded poset, with rk(x,y) = rk(x), but the Hasse diagram on the right represents a poset
v
i which does not have a grading.
X
WeobservethatwhenP hasanicestructure,E(P)commonlyhasanicestructureaswell. Inparticular
r
a we examine the boolean algebra of rank n, denoted B , which is defined to be the poset whose elements
n
are subsets of {1,...,n} with the relation given by containment, namely, for all x,y ∈ B , x≤ y if x is
n
a subset of y.
ThroughoutthepaperwesaythatagroupGactsonP ifitactsontheelementsofP andtheactionis
order-preservingandrank-preserving,thatis,forallg ∈Gwehavex≤y ⇔gx≤gy andrk(gx)=rk(x).
By Theorem 2.6 and the fact that B is unitary peck, if G is any action on B , then B /G is Peck. We
n n n
conjecture the following.
Conjecture 1.3. If G⊆Aut(B ), then E(B /G) is Peck.
n n
We prove this conjecture holds whenever the group action of G on B has the following property.
n
Definition 1.4. A group action of G on P is common cover transitive (CCT) if whenever x,y,z ∈ P
such that x⋖z, y⋖z, and y ∈Gx there exists some g ∈Stab(z) such that g·x=y.
Theorem 1.5. If a group action of G on B is CCT, then E(B /G) is Peck.
n n
1
2 DAVIDHEMMINGER,AARONLANDESMAN,ZIJIANYAO
7
(5,7) (6,7) (5,7) (6,7)
5 6
(3,5) (4,6) (3,5) (4,6)
3 4
(1,3) (2,3) (2,4) (1,3) (2,3) (2,4)
1 2
(0,1) (0,2) (0,1) (0,2)
0
P E(P) E(P) with relation ≤E′ (not graded)
Figure 1. Examples of E
A large number of groupactions onB have the CCT property. We first provethat some basic group
n
actions on B are CCT. Throughout the paper we let a subgroup G⊆S act on B by letting it act on
n n n
the elements within subsets of [n]:={1,...,n}, i.e. g·x={g·i: i ∈x} for all g ∈G, x ∈B . We also
n
embed the dihedral group D into S by letting it act as rotations and reflections on the vertices of an
2n n
n-gon.
Proposition 1.6. The following actions are CCT.
(1) The action of S on B ,
n n
(2) The action of D on B ,
2p p
(3) The action of D on B ,
4p 2p
where p is prime.
Wefurthershowthatcovertransitivityispreservedundersemidirectproducts,allowingustodescribe
several large families of CCT actions in Subsection 4.2.
Proposition 1.7. Let G⊆Aut(P), H⊳G, K ⊂G such that G=H⋊K. Suppose that the action of H
is on P is CCT and the action of K on P/H is CCT. Then the action of G on P is CCT.
Thepaperisorganizedasfollows. InSection2wecoverthe necessarybackgroundforposetsandPeck
posets. In Section 3 we show that E is well-defined and prove Theorem 1.5 regarding CCT actions along
with various other nice properties of E. Section 4 contains the proofs of Propositions 1.6 and 1.7 as well
as some examples of families of group actions shown to be CCT by these propositions. In Section 5, we
obtain a very different proof of [5, Theorem 1.1], in the case that r=1.
2. Background
In this section we review necessary background for this paper.
A graded poset P is a poset with a rank function rk: P →Z satisfying the following conditions.
≥0
(1) If x∈P and x⋖y, then rk(x)+1=rk(y),
(2) If x<y then rk(x)<rk(y)
We denote the ith rank of P by P = {x ∈ P: rk(x) = i}. Additionally, if for all x ∈ P, we have
i
0≤rk(x)≤n, and there exists x,y with rk(x)=0,rk(y)=n, we say that P is a gradedposetof rank n.
PECKNESS OF EDGE POSETS 3
Throughout the paper we write x≤ y to denote that x is less than or equal to y under the relation
P
defined on the poset P. When the poset is clear we omit the P and simply write x≤y.
Let P,Q ∈ P be two finite graded posets. A map f: P → Q is a morphism from P to Q if it is
rank-preserving and order preserving, in other words, for all x,y ∈P, x≤ y implies f(x)≤ f(y) and
P Q
rk(x) = rk(f(x)). We say that f is injective/surjective/bijective if it is an injection/surjection/bijection
from P to Q as sets.
Remark 2.1. Notethatwedonotrequiretheimplicationthatf(x)≤ f(y)impliesx≤ y inorderfor
Q P
f to be a morphism. In particular this means that a bijective morphism f need not be an isomorphism,
since it will not necessarily have a two-sided inverse.
In what follows, let P be a poset of rank n, and write p =|P |. If we have
i i
p ≤p ≤...≤p ≥p ≥...≥p
0 1 k k+1 n
for some 0 ≤ k ≤ n, then P is rank-unimodal. If p = p for all 1 ≤ i ≤ n, then P is rank-symmetric.
i n−i
An antichain inP is aset ofelements inP thatarepairwiseincomparable. If no antichainin P is larger
than the largest rank of P, then P is Sperner. More generally, P is k-Sperner if no union of k disjoint
antichains in P is larger than the union of the largestk ranks of P. We say that P is strongly Sperner if
it is k-Sperner for all 1≤k ≤n.
Definition 2.2. P is Peck if P is rank-symmetric, rank-unimodal, and strongly Sperner.
LetV(P)andV(P )bethecomplexvectorspaceswithbases{x:x∈P}and{x:x∈P }respectively.
i i
NotethatwewillfrequentlyabusenotationandwriteP andP forV(P)andV(P )whenwhatwemean
i i
is clear from context. In determining whether P is Peck, it is often useful to consider certain linear
transformations on V(P).
Definition 2.3. A linear map U: V(P) → V(P) is an order-raising operator if U(V(P )) = 0 and for
n
all 0≤i≤n−1, x∈P we have
i
U(x)= c y
X x,y
y⋗x
for some constants c ∈ C. We say that U is the Lefschetz map if all c on the right hand side are
x,y x,y
equal to 1.
We then have the following well-known characterizationof Peck posets.
Lemma 2.4 ([7],Lemma1.1). P is Peck if and only if there exists an order-raising operator U such that
for all 0≤i< n, the map Un−2i: V(P )→V(P ) is an isomorphism.
2 i n−i
Definition 2.5. Ifthe LefschetzmapsatisfiestheconditionforU inLemma2.4,thenP is unitary Peck.
NotethatagroupGactsonP iftheactiondefinesanembeddingG֒→Aut(P). Wedefinethequotient
poset P/G to be the poset whose elements are the orbits of G, with the relation O ≤ O′ if there exist
x∈O, x′ ∈O′ such that x≤ x′. We will use the following result in the paper.
P
Theorem 2.6 ([8], Theorem 1). If P is unitary Peck and G⊆Aut(P), then P/G is Peck.
3. The Edge Poset
In Subsection 3.1 we show that E as described in Definition 1.1 is well-defined and prove some useful
propertiesofE. InSubsection3.2weprovethatEsendsself-dualposetstoself-dualposets. InSubsection
3.3 we give several equivalent definitions for CCT actions, and Subsection 3.4 is devoted to the proof of
Theorem 1.5.
4 DAVIDHEMMINGER,AARONLANDESMAN,ZIJIANYAO
3.1. Functoriality of E and Group Actions. FirstweshowthatEiswell-definedinLemmas3.1,3.2,
and 3.3. After showing that E is well-defined we then define a natural G action on E(P) and define a
surjection E(P)/G→E(P/G) that will be important for the proof of Theorem 1.5.
When the poset P is clear, we will use ≤E and ⋖E to refer to ≤E(P) and ⋖E(P). Similarly, in Subsec-
tion 3.4, we define posets H(Bn), and will use ≤H and ⋖H in place of ≤H(Bn) and ⋖H(Bn).
Lemma 3.1. The relation ≤E defines a partial order on E(P).
Proof. We have that (x,y) ≤E (x,y) and that ≤E is transitive by definition. It remains to be shown
that ≤E is antisymmetric. Suppose (x,y) ≤E (x′,y′) and (x′,y′) ≤E (x,y). Then x ≤P x′ ≤P x and
y ≤ y′ ≤ y, so x=x′ and y =y′ by antisymmetry of ≤ , hence (x,y)=(x′,y′). (cid:3)
P P P
Lemma 3.2. For P a graded poset, the object E(P) is a graded poset.
Proof. To show E(P) is graded, we must show that (x,y)⋖E(x′,y′) =⇒ rk(x,y)+1= rk(x′,y′). This
fact follows immediately from the definition of ⋖E and the definition rkE(x,y)=rkP(x). (cid:3)
Lemma 3.3. Let f: P →Q be a morphism of finitegraded posets, and define a map E(f): E(P)→E(Q)
by E(f)(x,y)=(f(x),f(y)) for all (x,y)∈E(P). Then
(1) E(f) is a morphism of finite graded posets
(2) E(idP)=idE(P)
(3) If g: Q→R is a morphism of finite graded posets, then E(g◦f)=E(g)◦E(f).
Proof.
Part (1) First, E(f) is rank-preserving,since for all (x,y)∈E(P) we have
rkE(x,y)=rkP(x)=rkP(f(x))=rkE(E(f)(x,y)).
Suppose (x,y)⋖E(x′,y′). Then x⋖P x′, y⋖P y′, and since f is order-preserving,it follows that f(x)⋖P
f(x′), f(y)⋖P f(y′). Hence E(f)(x,y) ⋖E E(f)(x′,y′). Since ≤E is the transitive closure of ⋖E, we
similarly obtain E(f) is order-preservingand hence a morphism of finite graded posets.
Part (2) This is trivial.
Part (3) For all (x,y)∈E(P) we have
E(g◦f)(x,y)=(g(f(x)),g(f(y)))=(E(g)◦E(f))(x,y).
(cid:3)
Remark 3.4. ByLemmas 3.1,3.2,and3.3,the edgeposetconstructionEdefines anendofunctoronthe
category of finite graded posets with rank-preserving morphisms.
Given a group action of G on P, we can now easily define a natural group action of G on E(P) using
Lemma 3.3. For all g ∈ G we have that multiplication by g is an automorphism of P, so it follows that
E(g) is an automorphism of E(P) and furthermore that this gives a well-defined group action by Lemma
3.3.
Definition 3.5. Given a G-action on P, define a G-action on E(P) by g·(x,y)=E(g)(x,y)=(gx,gy).
We then have a well-defined quotient poset E(P)/G. It is natural to ask whether the operation of
quotienting out by G commutes with E, that is, whether E(P/G) ∼= E(P)/G. Unfortunately the two
posets are rarely isomorphic, but there is always a surjection E(P)/G → E(P/G), and this surjection is
also an injection precisely when the G-action on P is CCT, as will be shown in Lemma 3.14.
Proposition 3.6. The map q: E(P)/G → E(P/G) defined by q(G(x,y)) = (Gx,Gy) is a surjective
morphism.
Proof. Note that q is well defined because if (x′,y′)=g(x,y)=(g·x,g·y) for some g ∈G, then x′ ∈Gx
and y′ ∈Gy. Clearly q is rank-preservingand surjective, so it suffices to show that q is order-preserving.
Suppose that G(x,y)⋖E(P)/GG(w,z). Then there exist some (x0,y0)∈G(x,y), (w0,z0)∈G(w,z) such
that x0⋖P w0 andy0⋖P z0. We then havethat(Gx,Gy)⋖E(P/G)(Gw,Gz) by definition. Since ≤E(P/G)
is the transitive closure of ⋖E(P/G), q is order-preserving. (cid:3)
PECKNESS OF EDGE POSETS 5
3.2. The Opposite Functor and Self Dual Posets. Next, we introduce the notion of a dual poset,
given by applying the opposite functor, op, to a graded poset. We will show that op commutes with E.
This willimply that E(P) is self-dual if P is, which, in turn, willimply that E(B /G) is self-dual for any
n
group action of G on B .
n
Definition 3.7. Let P be the category of graded posets and let op: P → P be the opposite functor,
defined on posets as follows. For P a poset, the elements of Pop are the same as those of P with order
relation ≤Pop defined by x ≤Pop y ⇔ x ≥P y. Induced maps on morphisms are given as follows: for
P,Q gradedposets with f: P →Q, then fop: Pop →Qop is defined by fop(x)=f(x). The poset Pop is
called the dual poset of P. A poset P is self-dual if there is an isomorphism of posets P ∼=Pop.
Remark 3.8. Note that it is easy to check op: P → P is indeed a covariant functor. In more abstract
terms, ifwe view P as a category,Pop is the opposite category. Additionally, opasdefined in this wayis
actuallyaendofunctoronthe categoryofallfiniteposets,whichrestrictstoafunctoronthesubcategory
of graded posets.
Lemma 3.9. The functor op: P→P commutes with the functor E: P→P. That is, E(Pop)∼=E(P)op.
Proof. ObservethatE(Pop)iscanonicallyisomorphictoE(P)op,asgivenbythemorphismF: E(Pop)→
E(P)op,(x,y) 7→ (x,y). The inverse to F is given by G: E(P)op → E(Pop),(x,y) 7→ (x,y). These maps
are welldefined because E(P)op andE(Pop) are the same as sets, and it follows from the definitions that
these are both morphisms of graded posets, and so F defines an isomorphism. (cid:3)
Proposition 3.10. If P is a self-dual poset, then E(P) is also self-dual.
Proof. Since P is self-dual, there is an isomorphism f: P → Pop. By functoriality of E, as shown in
Lemma 3.3, we obtainthat E(f): E(P)→E(Pop) is an isomorphism. By Lemma 3.9, there is an isomor-
phism E(Pop) ∼= E(P)op. Then, letting F: E(Pop) → E(P)op,(x,y) 7→ (x,y) be the same isomorphism
defined in the proof of Lemma 3.9, the composition F ◦E(f): E(P) → E(P)op defines an isomorphism,
so E(P) is self-dual. (cid:3)
Example 3.11. While E(P) is commonly Peck when P is Peck, E(P) need not be Peck in general.
Furthermore, adding the condition that P be self-dual does not change this fact. In Figure 3.2 we give
anexampleofaposetP suchthatP isunitaryPeckandself-dual,butE(P) isnotrank-unimodal,hence
not Peck.
(4,6) (5,6) (5,7)
6 7
4 5
(2,4) (3,5)
2 3
(0,2) (0,3) (1,3)
0 1
P E(P)
Figure 2. P is self-dual and unitary Peck, but E(P) is not Peck
Remark 3.12. Wheneverthereisanactionψ: G×[n]→[n],weobtainaninducedactionφ:G×B →
n
B defined by
n
φ(g,{x ,...,x })={ψ(g,x ),...,ψ(g,x )}.
1 k 1 k
6 DAVIDHEMMINGER,AARONLANDESMAN,ZIJIANYAO
It is easy to see that any action φ: G×B → B arises in this way. That is, for any action φ of G on
n n
B there existsanactionψ ofGon[n]suchthat φ(g,{x ,...,x })={ψ(g,x ),...,ψ(g,x )}. This fact,
n 1 k 1 k
that all actions on B are induced by actions on [n] follows from the more general fact about atomistic
n
lattices. Recall, L is an atomistic lattice if there exists a minimum element of L and a subset of L called
“atoms” such that any other element of L can be expressed as a join of atoms. Then, for L an atomistic
lattice, any poset automorphism f: L → L restricts to an automorphism of the atoms of L, and f is
uniquelydeterminedbythisrestrictiontoatoms. InthecaseofB ,theatomsarepreciselythesingletons,
n
and hence can be identified with [n]. Whenever an actionψ of G on [n] is given, we refer to the action φ
defined above as the induced action on B .
n
Corollary 3.13. For any action φ: G × B → B , both E(B /G) and E(B )/G are self-dual. In
n n n n
particular, they are both rank-symmetric.
Proof. ByRemark3.12,anyactionφ: G×B →B isinducedbyanactionψ: G×[n]→[n]. Usingthis,
n n
observethatforanyφ,theposetB /Gisself-dual,asthereisanisomorphismf: B /G→(B /G)op,G·
n n n
x7→G·([n]\x). ThismapiswelldefinedonGorbitsbecauseeveryactiononB isinducedbyanaction
n
on [n]. Then, by Proposition 3.10, it follows that E(B /G) is self-dual.
n
It only remains to prove that E(B )/G is self-dual. However, from Proposition 3.10, E(B ) is self-
n n
dual, with the isomorphism given by E(f): E(B ) → E(Bop) ∼= E(B )op, that is, the map sending a
n n n
(x,y) 7→ ([n] \ y,[n] \ x). Once again, since the action on B is induced by an action on [n], this
n
isomorphism descends to an isomorphism E(f)G: E(B )/G → (E(B )/G)op, and so E(B )/G is self-
n n n
dual. (cid:3)
3.3. Equivalent Definitions of CCT Actions. We next give four equivalent definitions of CCT ac-
tions. The equivalence of (1) and (2) in the following Lemma 3.14 uses the notion of dual posets, while
the equivalence of (1),(3), and (4) use the fact that q: E(P)/G→E(P/G),G(x,y)7→(Gx,Gy) is always
a surjection.
Lemma 3.14. Let G be a group acting on a graded poset P. The following are equivalent:
(1) The action of G on P is CCT.
(2) Whenever x⋖y,x⋖z, and y ∈Gz, there exists some g ∈Stab(x) with gx=z.
(3) The map q: E(P)/G → E(P/G) defined by q(G(x,y)) = (Gx,Gy) is a bijective morphism (but
not necessarily an isomorphism).
(4) For all i there is an equality |(E(P)/G) |=|(E(P/G)) |.
i i
Proof. First, we show (1) ⇔ (2). There is an isomorphism f: B → Bop,x 7→ [n]\x. This defines a
n n
bijection between triples (x,y,z) with x⋖z,y ⋖z with x ∈ Gy, and triples (a,b,c) with c⋖a,c⋖b
with a ∈ Gb, given by (x,y,z) 7→ (f(x),f(y),f(z)). Furthermore, some g ∈ G satisfies g ∈ Stab(z) and
gx=y, if and only if it also satisfies g ∈Stab(f(z)) and g·f(x)=f(y). If an action is CCT, all triples
(x,y,z)satisfythese properties,andcondition(2)saysalltriples(a,b,c)satisfythese properties. So,the
above shows that (1) is equivalent to (2).
Second, we show (1) ⇔ (3). Observe that q is a bijection exactly when there do not exist distinct
orbits G(x,y) 6= G(x′,y′) with x′ ∈ Gx, y′ ∈ Gy. Fix (x,y),(x′,y′) ∈ E(P) such that x′ ∈ Gx and
y′ ∈Gy. Pick a g ∈G such that g·y′ =y. Then (g·x′,y)∈G(x′,y′), so G(x,y)=G(x′,y′) if and only
if there exists some g′ ∈G such that g′·x=g·x′ and g′·y =y. Hence q is a bijection if and only if the
G action is CCT.
Finally,wecheck(3)⇔(4). Againusing Proposition3.6,the morphismq is alwayssurjective. Sincea
morphismisalwaysrankpreserving,itmustmap(E(P)/G) surjectivelyonto(E(P/G)) . However,since
i i
the posets are finite, this surjection is a bijection if and only if the sets have the same cardinality. (cid:3)
Remark 3.15. While q is a bijection if and only if the action of G on P is CCT, it is not true that if
the action of G on P is CCT, then q is an isomorphism. For example, take G = D ⊂ S acting by
20 10
reflections androtationson{1,2,...,10}and considerthe induced actiononB . FromProposition1.6,
10
this actionisCCT.However,considerx={2,4},y={1,2,4},a={2,4,7},b={2,4,6,7}. Thenwemay
observe (x,y),(a,b)∈E(B10) and Gx<Ga,Gy <Gb, so (Gx,Gy) <E (Ga,Gb). However, it is not true
that G(x,y)<E G(a,b).
PECKNESS OF EDGE POSETS 7
3.4. Proof of Theorem 1.5. In this section we prove Theorem 1.5, which we recall here:
Theorem 1.5. If a group action of G on B is CCT, then E(B /G) is Peck.
n n
The proof is largely based on the following Lemma.
Lemma 3.16. Let P,Q two graded posets with a morphism f: P → Q that is a bijection (but not
necessarily an isomorphism). If P is Peck, then Q is Peck.
Proof. Let rk(P) = rk(Q) = n. Since P is Peck there exists an order-raising operator U such that
Un−2i: V(P )→V(P )isanisomorphism. Sincef isaposetmorphism,itfollowsthatthemapf◦U◦
i n−i
f−1 isanorder-raisingoperatoronQ. Wethenhavethatf◦Un−2i◦f−1 = f ◦U ◦f−1 n−2i : V(Q )→
i
(cid:0) (cid:1)
V(Q ) is an isomorphism since Un−2i: V(P )→V(P ) is an isomorphism and f is a bijection.
n−i i n−i
(cid:3)
By Lemma 3.16 and Proposition 3.6, in order to prove Theorem 1.5 it suffices to prove that E(B )/G
n
is Peck. One way to do this is to provethat E(B ) is unitary Peck and then apply Theorem2.6. In fact,
n
this approach generalizes to an arbitrary poset P.
Theorem 3.17. If the action of G on P is CCT and E(P) is unitary Peck, then E(P/G) is Peck.
Proof. Since the G-actionisCCT,there isabijectionq: E(P)/G→E(P/G)by Lemma3.14. Since E(P)
isunitaryPeckwehavethatE(P)/G isPeckbyTheorem2.6,henceE(P/G) isPeckby Lemma3.16. (cid:3)
WeprovethatE(B )isunitaryPeckforn>2in[4,Section8],butunfortunatelytheproofistechnical
n
andcomputational. NotethatbyTheorem3.17,thisimmediatelyimpliesTheorem1.5. Fortunatelythere
is a cleaner – albeit less direct – route to proving Theorem 1.5. In order to avoid showing that E(B )
n
is unitary Peck, we define a graded poset H(B ) for all n such that H(B ) is easily seen to be unitary
n n
Peck in Corollary 3.26. Furthermore, we define H(B ) such that a group action of G on B induces a
n n
groupactiononH(B )(Lemma3.23)andthereisalwaysabijectivemorphismf: H(B )/G→E(B )/G
n n n
(Lemma 3.24). By the above discussion, Theorem 1.5 readily follows.
Definition 3.18. For P a graded poset, define the graded poset H(P) as follows. Let the elements
(x,y)∈H(P)be pairs(x,y)∈P×P suchthatx⋖y. Define (x,y)⋖H(x′,y′)ifx⋖x′,y⋖y′ andx′ 6=y.
Then, define ≤H to be the transitive closure of ⋖H, and define rkH(x,y)=rkP(x).
Example 3.19. WegiveanexampleoftheposetH(B )inFigure3. ObservethatH(B )canbewritten
3 3
as a disjoint union of three copies of B . This is a single case of the more generalphenomenon provenin
2
Proposition 3.25.
Remark 3.20. Note that by definition (x,y)⋖H (x′,y′) precisely when (x,y)⋖E (x′,y′) and x′ 6= y,
hence (x,y)⋖H (x′,y′) ⇒ (x,y)⋖E (x′,y′). In other words, H(P) has the same elements as E(P) but
with a weaker partial order.
Lemma 3.21. For P a graded poset, the object H(P), as defined in Definition 3.18, is a graded poset.
Proof. This follows immediately from Remark 3.20 and the fact that E(P) is graded. (cid:3)
Remark 3.22. While E: P → P is a functor, H is not a functor. In particular, it is not possible to
define H(f) for f a morphism. This is illustrated in Figure 4. For example, suppose we took f: P →Q
defined by f(1) = 1,f(2) = f(3) = 2,f(4) = 3. It is easy to see that there is no possible morphism
H(f): H(P)→H(Q) because there are no morphisms H(P)→H(Q).
Given an action of a group G on P, we define an action of G on H(P) as we did for E(P) by again
defining g·(x,y) = (gx,gy) for all (x,y) ∈ P. We will then have a well-defined quotient poset H(P)/G
with the same elements as E(P)/G.
Lemma 3.23. The automorphism defined by g·(x,y) = (gx,gy) for all g ∈ G, (x,y) ∈ H(P) yields a
well-defined group action of G on H(P).
8 DAVIDHEMMINGER,AARONLANDESMAN,ZIJIANYAO
{1,2,3}
{1,2} {1,3} {2,3}
{1} {2} {3}
∅
B
3
({2,3},{1,2,3}) ({1,3},{1,2,3}) ({1,2},{1,2,3})
({2},{1,2}) ({3},{1,3}) ({1},{1,2}) ({3},{2,3}) ({1},{1,3}) ({2},{2,3})
(∅,{1}) (∅,{2}) (∅,{3})
H(B )
3
Figure 3. B and H(B )
3 3
4 3
3 4 2
2 3 2
1 2 1
1 1
P Q H(P) H(Q)
Figure 4.
Proof. Letg ∈G. Since ≤H is the transitiveclosureof⋖H it suffices toshowthat forall(x,y),(x′,y′)∈
H(P) we have (x,y) ⋖H (x′,y′) ⇔ g(x,y) ⋖H g(x′,y′). Since g is an automorphism of P, we have
x ≤ x′ ⇔ gx ≤ gx′, y ≤ y′ ⇔ gy ≤ gy′, and y 6= x′ ⇔ gy 6= gx′, so the result follows from the
P P P P
definition of ≤H. (cid:3)
Lemma 3.24. The map f: H(P)/G → E(P)/G defined by G(x,y) 7→ G(x,y) is a bijective morphism
for any group action of G on P.
Proof. The elements of H(P)/G,E(P)/G are the same by definition, so it suffices to show that f is a
morphism. Since f is clearlyrank-preserving,itsuffices to show f is order-preserving. This is immediate
from Remark 3.20. (cid:3)
PECKNESS OF EDGE POSETS 9
The remaining step in the proof of Theorem 1.5 is to show that H(B ) is unitary Peck, which we do
n
by generalizing Example 3.19 and showing that H(B ) is isomorphic to the disjoint union of boolean
n
algebras.
Proposition 3.25. H(B ) is isomorphic to n disjoint copies of B .
n n−1
Proof. Let the n disjoint copies of B be labeled B(i) , 1≤i≤n, with the elements of B(i) labeled
n−1 n−1 n−1
x(i), x⊆{1,...,n−1}. We will show that the map
n
f: H(B )−→ B(i)
n [ n−1
i=1
(x,x∪i)7−→x(i)
is an isomorphism. Suppose we have (x,y),(x′,y′)∈H(Bn) with (x,y)⋖H(x′,y′). Let j ∈[n] such that
y′ = y∪{j}, and let i ∈ [n] such that x′ = x∪{i}. If i 6= j, then x′ = y, contradicting the assumption
that (x,y)⋖H(x′,y′). Thus x′ =x∪{i} and y′ =y∪{i} for some i∈[n].
Converselywecaneasilycheckthatifi6∈y,then(x,y)⋖H(x∪{i},y∪{i}). Itfollowsthatforallsubsets
w ⊂ [n] such that |w| = 1, there is an isomorphism from the subposet of elements {(x,y): y\x = w}
to B defined by (x,y) 7→ (x\w,y \w). Furthermore if y \x 6= y′ \x′, then (x,y) and (x′,y′) are
n−1
incomparable,sothesesubposetsindexedbywarepairwisedisjoint,andH(B )isisomorphictoncopies
n
of B . (cid:3)
n
Corollary 3.26. H(B ) is unitary Peck for all n≥0.
n
Proof. This follows immediately from Proposition 3.25 and the fact that B is unitary Peck. Indeed,
n−1
B is shown to be unitary Peck, in [8, Theorem 2a] by noting that B = (B )k and that B is clearly
n k 1 1
unitary Peck. (cid:3)
Corollary 3.27. H(B )/G is Peck for any subgroup G⊂Aut(B ).
n n
Proof. This follows from Corollary 3.26 and Theorem 2.6. (cid:3)
The next corollary will not be particularly relevant in proving Theorem 1.5, but let us note it as an
aside.
Corollary 3.28. Both E(B ) and H(B ) have symmetric chain decompositions (SCD).
n n
Proof. H(B ) has an SCD by Proposition 3.25 and the fact that B has an SCD, as shown in [3]. By
n n−1
Lemma 3.24 there is a bijective morphism f: H(B )→E(B ), and since a bijective morphism takes an
n n
SCD to an SCD it follows that E(B ) has an SCD. (cid:3)
n
Corollary 3.29. E(B )/G is Peck for any subgroup G⊂Aut(B ).
n n
Proof. ByCorollary3.27,H(B )/GisPeck. ByLemma3.24,themapf: H(B )/G→E(B )/G,G(x,y)7→
n n n
G(x,y) is a bijective morphism. Then, by Lemma 3.16, it follows that E(B )/G is Peck. (cid:3)
n
We now deduce Theorem 1.5.
Proof of Theorem 1.5. By Corollary 3.29, E(B )/G is Peck for any group action of G on B . Since
n n
the G-action is CCT, there is a bijective morphism from E(B )/G to E(B /G) by Lemma 3.14. Hence
n n
E(B /G) is Peck by Lemma 3.16. (cid:3)
n
NotethatwehavealsodevelopedseveralgeneralizationsofE,forwhichmanysimilarresultshold. For
more information, see [[4], Subsection 3.3]
10 DAVIDHEMMINGER,AARONLANDESMAN,ZIJIANYAO
4. Common Cover Transitive Actions
In this section, we develop the theory of CCT actions φ where G is a group, P is a poset, and
φ:G×P →P isanaction. RecallDefinition1.4,thatφisCCTifwheneverx,y,z ∈P,x⋖z,y⋖z,x∈Gy
then there exists g ∈ Stab(z) with gx = y. We show that the CCT property is closed under semidirect
products,intheappropriatesense. FromProposition1.6,whichwillbeproveninSubsubsection4.2.4,the
actionofS onB andtheactionofcertaindihedralgroupsareCCT.We canthenusethese asbuilding
n n
blocks to construct other CCT groups. In particular, we shall show in this section that automorphism
groups of rooted trees are CCT.
Example 4.1. Two rather trivialexamples ofCCT actions are φ: S ×B →B and ψ: G×B →B
n n n n n
where G is arbitrary, φ is the action induced by S permuting the elements of [n], and ψ is the trivial
n
action. In the formercase, E(B /S )is simply a chainwith npoints, andsois E(B )/S , since all(x,y)
n n n n
are identified under the S action. In the latter case,since Gacts trivially by φ, both E(B /G)∼=E(B )
n n n
and E(B )/G∼=E(B ). So again, ψ is CCT.
n n
4.1. Preservation Under Semidirect Products.
Lemma 4.2. Let G⊆Aut(P), H⊳G, K ⊂G such that G=H⋊K. We then have a well defined group
action K×P/H →P/H,(k,Hx)7→H(k·x).
Proof. Note that if x,x′ ∈Hx, we have x′ =h·x for some h∈H. Since H is normalin G, we have that
for all k ∈G there exists h′ ∈H so that khk−1 =h′. So
k·x′ =kh·x=k(k−1h′k)·x=h′·(k·x)
Hence k·x and k·x′ are in the same H-orbit, so we have a well-defined group action of K on P/H
defined by k·Hx=H(k·x). (cid:3)
Recall Proposition 1.7, as stated in the introduction, which says that the CCT property is preserved
undersemidirectproducts. WewilluseProposition1.7toconstructmoreexamplesofCCTgroupactions,
in particular using it to give a simple proofs that CCT actions are preserved under direct products and
wreath products.
Proposition 1.7. Let G⊆Aut(P), H⊳G, K ⊂G such that G=H⋊K. Suppose that the action of H
is on P is CCT and the action of K on P/H is CCT. Then the action of G on P is CCT.
Proof. Since G=H⋊K, every element g ∈G can be written uniquely as a product hk for some h∈H,
k ∈ K. Let x,y,z ∈ P such that x⋖z, y ⋖z and such that there exists some h k ∈ G such that
0 0
h k ·x=y. It suffices to show that there exists some g ∈Stab(z) such that g·x=y.
0 0
TheorbitsHx,Hy,Hz ∈P/H satisfyHx⋖Hz,Hy⋖Hz suchthatk ·Hx=Hy,sosincethe action
0
of K on P/H is CCT there exists some k ∈ K such that k ∈ Stab(Hz) and k ·Hx = Hy. It follows
1 1 1
that there exists some h ∈H such that h k h ∈Stab(z) and h k h ·x∈Hy.
1 1 1 0 1 1 0
Write x′ = h k h ·x. Since the group action of G must be order-preserving by definition, we have
1 1 0
that x′ ⋖z. We already had that y⋖z and x′ ∈ Hy, hence there exists some h ∈ Stab(z) such that
2
h ·x′ = y by the fact that the action of H on P is CCT. Then we have that h h k h ·x= h ·x′ =y
2 2 1 1 0 2
and h h k h ·z =h ·z =z, as desired. (cid:3)
2 1 1 0 2
Proposition 4.3. For φ: G×P →P,ψ: H ×Q→Q two CCT actions, then the direct product
φ×ψ: (G×H)×(P ×Q)→(P ×Q),(g,h)·(x,y)7→(gx,hy)
is also CCT.
Proof. FirstnotethatifeitherGorH actstrivially,thenitcanbeeasilycheckedthattheactionofG×H
is CCT. Next, observe that G×H can be viewed as the semidirect product of (G×{e})⋊({e}×H).
Since the action of G on P is CCT, the action of G×{e} on P ×Q is CCT. Also, since the action of
H on Q is CCT, it follows that the action of {e}×H on (P/G)×Q is CCT. Therefore, the action of
(G×{e})⋊({e}×H) satisfies the conditions of Proposition1.7 and so the action of G×H is CCT. (cid:3)